Chapter 5.5, Problem 127P

To determine

The final temperature in the balloon.

Answer to Problem 127P

The final temperature in the balloon is $315\text{K}$.

Explanation of Solution

Write the equation of mass balance.

$m_{\text{in}}-m_{\text{e}}=\Delta m_{\text{system}}$ (I)

Here, the inlet mass is $m_{\text{in}}$, the exit mass is $m_{\text{e}}$ and the change in mass of the system is $\Delta m_{\text{system}}$.

The change in mass of the system for the control volume is expressed as,

$\Delta m_{\text{system}}={\left(m_2-m_1\right)}_{\text{cv}}$

Here, the suffixes 1 and 2 indicates the initial and final states of the system.

Consider the given balloon as the control volume. Initially the balloon is filled with helium and connected to the supply line with the valve, the valve is opened, and helium is allowed to enter the balloon until it reaches the pressure of supply line. No mass is allowed to exit the balloon i.e. $m_{\text{e}}=0$.

Rewrite the Equation (I) as follows.

$\begin{array}{l}{m}_{\text{in}}-0={\left(m_2-m_1\right)}_{\text{cv}}\\ m_{\text{in}}=m_2-m_1\end{array}$ (II)

Write the formula for initial and final masses.

$m_{\text{1}}=\frac{P_1{\nu }_1}{R{T}_1}$ (III)

$m_{\text{2}}=\frac{P_2{\nu }_2}{R{T}_2}$ (IV)

Here, the pressure is $P$, the volume is $\nu$, the gas constant of is $R$, the temperature is $T$ and the subscripts 1 and 2 indicates the initial and final states.

Write the pressure and volume relation as follows.

$\frac{P_1}{P_2}=\frac{{\nu }_1}{{\nu }_2}$ (V)

Write the energy balance equation.

$\begin{array}{l}{E}_{in}-E_{out}=\Delta E_{system}\\ \left\{\begin{array}{l}\left[Q_{\text{in}}+W_{\text{in}}+m_{\text{in}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{in}}\right]\\ -\left[Q_{\text{e}}+W_{\text{e}}+m_{\text{e}}{\left(h+\text{ke}+\text{pe}\right)}_{\text{e}}\right]\end{array}\right\}={\left[\begin{array}{l}{m}_2{\left(u+\text{ke}+\text{pe}\right)}_2\\ -m_1{\left(u+\text{ke}+\text{pe}\right)}_1\end{array}\right]}_{\text{system}}\end{array}$ (VI)

Here, the heat transfer is $Q$, the work transfer is $W$, the enthalpy is $h$, the internal energy is $u$, the kinetic energy is $\text{ke}$, the potential energy is $\text{pe}$ and the change in net energy of the system is $\Delta E_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The balloon expands when further helium is filled and the boundary work is done. i.e. $W_{\text{in}}=0,W_{\text{e}}=W_{\text{b,out}}$. There is no heat transfer .i.e. $Q_{\text{in}}=Q_{\text{e}}=0$. Neglect the kinetic and potential energy changes i.e. $\left(\Delta \text{ke}=\Delta \text{pe}=0\right)$.

The Equation (VI) reduced as follows.

$\begin{array}{l}{m}_{\text{in}}{h}_{\text{in}}-W_{\text{b,out}}=m_2{u}_2-m_1{u}_1\\ W_{\text{b,out}}=m_{\text{in}}{h}_{\text{in}}-m_2{u}_2+m_1{u}_1\end{array}$ (VII)

Write the general formula for boundary work by the helium (expansion of balloon).

$\begin{array}{c}{W}_{\text{b,out}}=P\Delta \nu \\ =\left(\frac{P_1+P_2}{2}\right)\left({\nu }_2+{\nu }_1\right)\end{array}$ (VIII)

Write the general expressions for enthalpy and internal energy.

$\begin{array}{l}h=c_{p}T\\ u=c_{v}T\end{array}$

Here, the specific heat at constant pressure is $c_p$, the specific heat at constant volume is $c_v$ and the temperature is $T$.

Rewrite the equation (VII) as follows with reference to the general expression of enthalpy and internal energy.

$W_{\text{b,out}}=m_{\text{in}}{c}_p{T}_{\text{in}}-m_2{c}_v{T}_2+m_1{c}_v{T}_1$ (IX)

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The gas constant $\left(R\right)$ of helium is $2.0769\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$.

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The specific heat at constant pressure $\left(c_p\right)$ is $5.1926\text{kJ/kg}\cdot \text{K}$ and the specific heat at constant volume $\left(c_v\right)$ is $3.1156\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $100\text{kPa}$ for $P_1$, $40{\text{m}}^3$ for ${\nu }_1$, $2.0769\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$, and $17°\text{C}$ for $T_1$ in Equation (III).

$\begin{array}{c}{m}_{\text{1}}=\frac{\left(100\text{kPa}\right)\left(40{\text{m}}^3\right)}{\left(2.0769\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)\left(17°\text{C}\right)}\\ =\frac{\left(100\text{kPa}\right)\left(40{\text{m}}^3\right)}{\left(2.0769\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)\left(17+273\right)\text{K}}\\ =\frac{4000\text{kPa}{\text{.m}}^3}{602.301\text{kPa}{\text{.m}}^3\text{/kg}}\\ =6.6412\text{kg}\end{array}$

Substitute $100\text{kPa}$ for $P_1$, $125\text{kPa}$ for $P_2$, $40{\text{m}}^3$ for ${\nu }_1$ in Equation (V)

$\begin{array}{l}\frac{100\text{kPa}}{125\text{kPa}}=\frac{40{\text{m}}^3}{{\nu }_2}\\ {\nu }_2=\frac{\left(40{\text{m}}^3\right)\left(125\text{kPa}\right)}{100\text{kPa}}\\ =50{\text{m}}^3\end{array}$

Substitute $125\text{kPa}$ for $P_2$, $50{\text{m}}^3$ for ${\nu }_2$ and $2.0769\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}$ for $R$ in Equation (IV).

$\begin{array}{c}{m}_{\text{2}}=\frac{\left(125\text{kPa}\right)\left(50{\text{m}}^3\right)}{\left(2.0769\text{kPa}{\text{.m}}^3\text{/kg}\cdot \text{K}\right)T_2}\\ =\frac{3009.2927}{T_2}\text{kg}\end{array}$

Substitute $\frac{3009.2927}{T_2}\text{kg}$ for $m_2$ and $6.6412\text{kg}$ for $m_1$ in Equation (II).

$m_{\text{in}}=\frac{3009.2927}{T_2}\text{kg}-6.6412\text{kg}$

Substitute $100\text{kPa}$ for $P_1$, $125\text{kPa}$ for $P_2$, $40{\text{m}}^3$ for ${\nu }_1$ and $50{\text{m}}^3$ for ${\nu }_2$ in Equation (VIII).

$\begin{array}{c}{W}_{\text{b,out}}=\left(\frac{100\text{}\text{kPa}+125\text{kPa}}{2}\right)\left(50{\text{m}}^3-40{\text{m}}^3\right)\\ =1125\text{kPa}\cdot {\text{m}}^3\\ =1125\text{kPa}\cdot {\text{m}}^3\times \frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^3}\\ =1125\text{kJ}\end{array}$

Substitute $1125\text{kJ}$ for $W_{\text{b,out}}$, $\frac{3009.2927}{T_2}\text{kg}-6.6412\text{kg}$ for $m_{\text{in}}$, $5.1926\text{kJ/kg}\cdot \text{K}$ for $c_p$, $25°\text{C}$ for $T_{\text{in}}$, $\frac{3009.2927}{T_2}\text{kg}$ for $m_2$, $3.1156\text{kJ/kg}\cdot \text{K}$ for $c_v$, $6.6412\text{kg}$ for $m_1$, $3.1156\text{kJ/kg}\cdot \text{K}$ for $c_v$, and $17\text{°C}$ for $T_1$, in Equation (IX).

$\begin{array}{l}1125\text{kJ}=\left[\begin{array}{l}\left(\frac{3009.2927}{T_2}\text{kg}-6.6412\text{kg}\right)\left(5.1926\text{kJ/kg}\cdot \text{K}\right)\left(25°\text{C}\right)\\ -\left(\frac{3009.2927}{T_2}\text{kg}\right)\left(3.1156\text{kJ/kg}\cdot \text{K}\right)T_2\\ +\left(6.6412\text{kg}\right)\left(3.1156\text{kJ/kg}\cdot \text{K}\right)\left(17\text{°C}\right)\end{array}\right]\\ 1125\text{kJ}=\left[\begin{array}{l}\left(\frac{3009.2927}{T_2}\text{kg}-6.6412\text{kg}\right)\left(5.1926\text{kJ/kg}\cdot \text{K}\right)\left(25+273\right)\text{K}\\ -\left(\frac{3009.2927}{T_2}\text{kg}\right)\left(3.1156\text{kJ/kg}\cdot \text{K}\right)T_2\\ +\left(6.6412\text{kg}\right)\left(3.1156\text{kJ/kg}\cdot \text{K}\right)\left(17+273\right)\text{K}\end{array}\right]\\ 1125\text{kJ}=\left[\begin{array}{l}\left(\frac{3009.2927}{T_2}\text{kg}-6.6412\text{kg}\right)\left(1547.3948\text{kJ/kg}\right)\\ -9375.7523\text{kJ}+6000.4836\text{kJ}\end{array}\right]\\ 1125\text{kJ}=\left[\begin{array}{l}\left(\frac{3009.2927}{T_2}\left(1547.3948\text{kJ}\right)\right)-\left(6.6412\text{kg}\right)\left(1547.3948\text{kJ/kg}\right)\\ -3375.2687\text{kJ}\end{array}\right]\end{array}$

$\begin{array}{l}1125\text{kJ}=\left[\begin{array}{l}\left(\frac{3009.2927}{T_2}\left(1547.3948\text{kJ}\right)\right)-10276.5584\text{kJ}\\ -3375.2687\text{kJ}\end{array}\right]\\ 1125\text{kJ}=\frac{3009.2927}{T_2}\left(1547.3948\text{kJ}\right)-13651.8271\text{kJ}\\ 1125\text{kJ}+13651.8271\text{kJ}=\frac{3009.2927}{T_2}\left(1547.3948\text{kJ}\right)\\ \frac{14776.8271\text{kJ}}{1547.3948\text{}\text{kJ}}=\frac{3009.2927}{T_2}\end{array}$

$\begin{array}{l}9.5495=\frac{3009.2927}{T_2}\\ T_2=\frac{3009.2927}{9.5495}\\ =315.1261\text{K}\\ \simeq 315\text{K}\end{array}$

Thus, the final temperature in the balloon is $315\text{K}$.